Coffee Cooling Project
(1) Object:
To determine the cooling rate of coffee in 3 sizes of lidded coffee cups (A&W small, medium, large) in single and double-cupped cases, and compare and analyze these results.
(2) Materials:
TI-83 Calculator (CHEM/BIO APPS Program)
Temperature Probe
CBL System
Hot Coffee
Lidded Coffee Cups
(3) Theory:
Isaac Newton determined the cooling process between one constant, the initial difference in temperature between coffee and room temperature, (To), and the changing temperature difference between the coffee and room temperature, (T), to be an exponential decay process that could be described by this formula:
T=To(e)-kt → T= temperature difference between coffee and room temperature (ºC)
→To= temperature difference between coffee and room temperature (ºC) at t=0
→k= decay rate (ºC/s) based on insulation of the cup
time to cool a DT
Rate of Cooling: dT/dt= -kT
The following is additional information from the experiment of others whom determined the results of adding cream to the coffee while it cools.
The results should turn out as follows: The final temperature of the coffee with cream added early (Part A) is a few degrees above the coffee with cream added later (Part B). If the room temperature is 24 or 25 C, there will be little difference. The experiment works best if the temperature of the room is about 20 C, as in the wintertime. Also, the graph for Part A after the cream is added is more horizontal, or has a lower slope, than the graph for Part B until the cream is added. That shows that if the temperature between the liquid (coffee) and the room is greater, the cooling is faster and if the difference in temperatures is less, the cooling is slower. Or, the rate is greater is the temperature difference is greater. As was mentioned before,
Time to cool α ΔT (Newton’s Law of Cooling).
So, add the cream in the beginning to enjoy your coffee as hot as possible after fifteen minutes.
Additional Information is from:
http://www.haverford.edu/educ/knight-booklet/newtoncool.htm
(4) Procedure:
See Attached Page for Procedure.
(5) Data:
See Attached Page for Data.
(6) Calculations:
1. Determine the exponential best fit curve for L1, L3 for each cup type in the form T=a(b)t.
Single Cup
/Double Cup
Small / T= 53.8(0.999732)t / T= 55.1(0.999718)tMedium / T= 55.4(0.999752)t / T= 55.5(0.999736)t
Large / T= 55.0(0.999766)t / T= 56.0(0.999744)t
2. Determine each equation in base “e”, by calculating ln(b), and then writing it in the form T=a(e)kt, where k is the cooling rate constant, where t is in seconds and then minutes.
Single Cup
/Double Cup
Small / T= 53.8(e)-2.68331598X10^-4 tT= 53.8(e)-0.016099998959 t / T= 55.1(e)-2.81779196X10^-4 t
T= 55.1(e)-0.0169067518 t
Medium / T= 55.4(e)-2.47586447X10-4 t
T= 55.4(e)-0.01148551868 t / T= 55.5(e)-2.63510716X10^-4 t
T= 55.5(e)-0.015810643 t
Large / T= 55.0(e)-2.34166015X10^-4 t
T= 55.0(e)-0.0140499609 t / T= 56.0(e)-2.55564454X10^-4 t
T= 56.0(e)-0.0153338672 t
3. Determine the first derivative (rate of cooling) of each cup in ºC/s and ºC/min.
Single Cup
/Double Cup
Small / dT/dt= -2.68331598X10-4 TdT/dt= -0.016099998959 T / dT/dt= -2.81779196X10-4 T
dT/dt= -0.0169067518 T
Medium / dT/dt= -2.47586447X10-4 T
dT/dt= -0.01148551868 T / dT/dt= -2.63510716X10-4 T
dT/dt= -0.015810643 T
Large / dT/dt= -2.34166015X10-4 T
dT/dt= -0.0140499609 T / dT/dt= -2.55564454X10-4 T
dT/dt= -0.0153338672 T
4. For each cup type determine the rate of cooling (ºC/min) when the temperature difference between the coffee and air (T) is:
T / dT/dt/ / Small / Medium / Large
Single / Double / Single / Double / Single / Double
55ºC / -0.885 / -0.930 / -0.817 / -0.870 / -0.773 / -0.843
45ºC / -0.724 / -0.761 / -0.668 / -0.711 / -0.632 / -0.690
35ºC / -0.563 / -0.592 / -0.520 / -0.553 / -0.492 / -0.537
25ºC / -0.402 / -0.423 / -0.371 / -0.395 / -0.351 / -0.383
5. Use your equation to determine the time (to the nearest 0.1 min) it would take for the coffee to reach a temperature of 1ºC above room temperature (T= 1.0).
Single Double
T= 55.0(e)-0.0140499609 t T= 56.0(e)-0.0153338672 t
1.0= 55.0(e)-0.0140499609 t 1.0= 56.0(e)-0.0153338672 t
ln(1.0/55.0) = -0.0140499609 t ln(1.0/56.0) = -0.0153338672 t
t = 285.2 min t = 262.5 min
(7) Analysis:
1. What is the value of “k” and how does it vary in this experiment?
In this experiment, the value of “k” represents the cooling rate constant of the coffee. “k” is a negative value because it represents the cooling of the coffee, thus a decrease in temperature, so therefore it is negative. The larger the value of “k” is, the faster the coffee cooled, and the smaller the value of “k” is, the slower the coffee cooled.
For the large single cup, k= -0.0140499609
For the large double cup, k= -0.0153338672
For the experiment, the “k” values varied from the largest being the small double cup, to the smallest being the large single cup.
2. Compare the cooling rates between small, medium, and large cups at the temperature differences calculated.
a) at the temperature differences calculated, and in all cases, the small cup cooled the fastest, followed by the medium cup, with the large cup cooling the slowest
b) at the temperature differences calculated, and in all cases, the double cup cooled faster than the single cup
3. What factors may have affected the experiment?
· The coffee being poured twice at two different times
· The amount and regularity of stirring the coffee
· The air conditioning may have created a draft on the double cup side of the table
· The precise amount of coffee poured into each cup
· The fact that there was two different pots of coffee used for the experiment
4. What variables did we try to keep constant?
· Amount of coffee was about ¾” from the rim
· The room temperature
· The same lid for each cup
· Each cup was made from the same material
· Each probe was poked through a hole in the center of the lid
(8) Summary:
· The conclusive evidence from this experiment suggests that the larger the size of the coffee cup, the longer it retained its heat and kept the coffee warm. The large cup retained its heat the best, followed by the medium cup, and rounded off by the small cup. From the data collected in this experiment, there was not a significant difference in the loss of temperature between the single cup, and the double cup, although the single cup slightly performed better overall, in all cup sizes, although there is a possible explanation for this stated in the analysis. If taken theoretically, the data from this experiment may suggest that a larger cup of pop would stay cooler, longer, than a smaller one of similar materials, and lid.
· To better retain its heat, the lid is the most likely part of the cup that needs to be changed, because heat rises, and the lid is the upper most part of the cup that would trap the heat. A lid consisting of similar cardboard-like materials as the rest of the cup with a similar shape and size as the plastic lids would be a better approach. A thin wire ring, or some sort of elastic would be used to secure the lid under the lip of the cup, or if production permits, a similar groove as on the plastic lids that fits on the lip of the cup would also work. The upper portion of the lid would be like a flattened out dome. On one side there would be a small pinhole to allow for airflow when drinking, and on the other side would be the area for drinking. There would be a hole for drinking out of, but it would be covered with a flap of similar material as the cup and lid. Toward the center of the cup, and the pinhole, the flap would be attached to the lid with a permanent adhesive. Surrounding the hole for drinking out of, the flap would be attached with a temporary adhesive that could be removed and then re-covered by the customer. The flap would extend over the edge of the lid for ease of lifting it. The idea is that the coffee would be virtually completely sealed within the cup, and the only time that excessive quantities of heat would be allowed to escape is when the customer would lift the flap to take a drink.